**Maharashtra state Board 10th Standard Solutions Chapter 1 Linear Equations in two Variables** – Here are all the MH Board solutions for 10th Standard maths Problem Set 1. This solution contains questions, answers, images, explanations of the complete Problem Set 1 titled Linear Equations in two Variables of maths taught in 10th Standard. If you are a student of 10th Standard who is using Maharashtra state Board Textbook to study maths, then you must come across Problem Set 1 **Linear Equations in two Variables**. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete Maharashtra Board Solutions for 10th Standard maths Chapter 1 Linear Equations in two Variables in one place.

## Maharashtra State Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Problem Set 1

Choose correct alternative for each of the following questions.

Question 1.

To draw graph of 4x + 5y = 19, find y when x = 1.

(a) 4

(b) 3

(c) 2

(d) -3

Answer:

(b)

Question 2.

For simultaneous equations in variables x and y, D_{x} = 49, Dy = – 63, D = 7 then what is x?

(a) 7

(b) -7

(c) \(\frac { 1 }{ 7 } \)

(d) \(\frac { -1 }{ 7 } \)

Answer:

(a)

Question 3.

Find the value of

(a) -1

(b) -41

(c) 41

(d) 1

Answer:

(d)

Question 4.

To solvex + y = 3; 3x – 2y – 4 = 0 by determinant method find D.

(a) 5

(b) 1

(c) -5

(d) -1

Answer:

(c)

Question 5.

ax + by = c and mx + n y = d and an ≠ bm then these simultaneous equations have-

(a) Only one common solution

(b) No solution

(c) Infinite number of solutions

(d) Only two solutions.

Answer:

(a)

Question 2.

Complete the following table to draw the graph of 2x – 6y = 3.

Answer:

Question 3.

Solve the following simultaneous equations graphically.

i. 2x + 3y = 12 ; x – y = 1

ii. x – 3y = 1 ; 3x – 2y + 4 = 0

iii. 5x – 6y + 30 = 0; 5x + 4y – 20 = 0

iv. 3x – y – 2 = 0 ; 2x + y = 8

v. 3x + y= 10 ; x – y = 2

Answer:

i. The given simultaneous equations are

The two lines interest at point (3,2).

∴ x = 3 and y = 2 is the solution of the simultaneous equations 2x + 3y = 12 and x – y = 1.

ii. The given simultaneous equations are

The two lines intersect at point (-2, -1).

∴ x = -2 and y = -1 is the solution of the simultaneous equations x – 3y = 1 and 3x – 2p + 4 = 0.

iii. The given simultaneous equations are

The two lines intersect at point (0, 5).

∴ x = 0 and y = 5 is the solution of the simultaneous equations 5x – 6y + 30 = 0 and 5x + 4y – 20 = 0.

iv. The given simultaneous equations are

The two lines intersect at point (2, 4).

∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.

v. The given simultaneous equations are

The two lines intersect at point (3, 1).

∴ x = 3 and y = 1 is the solution of the simultaneous equations 3x + y = 10 and x – y = 2.

Question 4.

Find the values of each of the following determinants.

Solution:

Question 5.

Solve the following equations by Cramer’s method.

Solution:

i. The given simultaneous equations are

6x – 3y = -10 …(i)

3x + 5y – 8 = 0

∴ 3x + 5y = 8 …(ii)

Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with a_{1}x + b_{1}y = c_{1} and a_{2}x + b_{2}y = c_{2}, we get

a_{1} = 6, b_{1} = -3, c_{1} = 10 and

a_{2} = 3, b_{2} = 5, c_{2} = 8

ii. The given simultaneous equations are

4m – 2n = -4 …(i)

4m + 3n = 16 …(ii)

Equations (i) and (ii) are in am + bn = c form.

Comparing the given equations with a_{1}m + b_{1}n = c_{1} and a_{2}m + b_{2}n = c_{2}, we get

a_{1} = 4, b_{1} = -2, c_{1} = -4 and

a_{2} = 4, b_{2} = 3, c_{2} = 16

∴ (m, n) = (1, 4) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

iv. The given simultaneous equations are

7x + 3y = 15 …(i)

12y – 5x = 39

i.e. -5x + 12y = 39 …(ii)

Equations (i) and (ii) are in ax + by = c form.

Comparing the given equations with

a_{1}x + b_{1}y = c_{1} and a_{2}x + b_{2}y = c_{2}, we get

a_{1} = 7, b_{1} = 3, c_{1} = 15 and

a_{2} = -5, b_{2} = 12, c_{2} = 39

v. The given simultaneous equations are

∴ 4(x + y – 8) = 2(3x – y)

∴ 4x + 4y – 32 = 6x – 2y

∴ 6x – 4x – 2y – 4y = -32

∴ 2x – 6y = -32

∴ x – 3y = -16 …(ii)[Dividing both sides by 2]

Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with

a_{1}x + b_{1}y = c_{1} and a_{2}x + b_{2}y = c_{2}, we get

a_{1} = 1, b_{1} = -1, c_{1} = -4 and

a_{2} = 1, b_{2} = -3, c_{2} = -16

∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.

Question 6.

Solve the following simultaneous equations:

Answer:

i. The given simultaneous equations are

Subtracting equation (iv) from (iii), we get

∴ (x, y) = (6, – 4) is the solution of the given simultaneous equations.

ii. The given simultaneous equations are

Adding equations (v) and (vi), we get

iii. The given simultaneous equations are

∴ (x, y) = (1, 2) is the solution of the given simultaneous equations.

iv. The given simultaneous equations are

∴ Equations (i) and (ii) become 7q – 2p = 5 …(iii)

8q + 7p = 15 …(iv)

Multiplying equation (iii) by 7, we get

49q – 14p = 35 …(v)

Multiplying equation (iv) by 2, we get

16q + 14p = 30 …(vi)

Adding equations (v) and (vi), we get

Substituting q = 1 in equation (iv), we get

8(1) + 7p = 15

∴ 8 + 7p = 15

∴ 7p = 15 – 8

∴ 7p = 7

∴ p = \(\frac { 7 }{ 7 } \) = 1

∴ (P, q) = (1,1)

Resubstituting the values of p and q, we get

1 = \(\frac { 1 }{ x } \) and 1 = \(\frac { 1 }{ y } \)

∴ x = 1 and y = 1

∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.

v. The given simultaneous equations are

Resubstituting the values of p and q, we get

∴ 3x + 4y = 10 …(v)

and 2x – 3y = 1 …(vi)

Multiplying equation (v) by 3, we get

9x + 12y = 30 …(vii)

Multiplying equation (vi) by 4, we get

8x – 12y = 4 …(viii)

Adding equations (vii) and (viii), we get

Substituting x = 2 in equation (v), we get

3(2) + 4y = 10

⇒ 6 + 4y = 10

⇒ 4y = 10 – 6

⇒ y = 4/4 = 1

∴ y = 1

∴ (x, y) = (2, 1) is the solution of the given simultaneous equations.

Question 7.

Solve the following word problems, i. A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.

Solution:

Let the digit in unit’s place be x

and that in the ten’s place be y.

ii. Kantabai bought 1 \(\frac { 1 }{ 2 } \) kg tea and 5 kg sugar from a shop. She paid ₹ 50 as return fare for rickshaw. Total expense was ₹ 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So, next month she placed the order online for 2 kg tea and 7 kg sugar. She paid ₹ 880 for that. Find the rate of sugar and tea per kg.

Solution:

Let the rate of tea be ₹ x per kg and that of sugar be ₹ y per kg.

According to the first condition,

cost of 1 \(\frac { 1 }{ 2 } \) kg tea + cost of 5 kg sugar + fare for rickshaw = total expense

According to the second condition,

cost of 2 kg tea + cost of 7 kg sugar = total expense

2x + 7y = 880 …(ii)

Multiplying equation (i) by 2, we get

6x + 20y = 2600 …(iii)

Multiplying equation (ii) by 3, we get

6x + 21y = 2640 …(iv)

Subtracting equation (iii) from (iv), we get

∴ The rate of tea is ₹ 300 per kg and that of sugar is ₹ 40 per kg.

iii. To find number of notes that Anushka had, complete the following activity.

Solution:

Anushka had x notes of ₹ 100 and y notes of ₹ 50.

According to the first condition,

100x + 50y = 2500

∴ 2x + y = 50 …(i) [Dividing both sides by 50]

According to the second condition,

100y + 50x = 2000

∴ 2y + x = 40 … [Dividing both sides by 50]

i.e. x + 2y = 40

∴ 2x + 4y = 80 …(ii) [Multiplying both sides by 2]

Subtracting equation (i) from (ii), we get

∴ Anushka had 20 notes of ₹ 100 and 10 notes of ₹ 50.

iv. Sum of the present ages of Manish and Savita is 31, Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.

Solution:

Let the present ages of Manish and Savita be x years and y years respectively.

According to the first condition,

x + y = 31 …(i)

3 years ago,

Manish’s age = (x – 3) years

Savita’s age = (y – 3) years

According to the second condition,

(x – 3) = 4 (y – 3)

∴ x – 3 = 4y – 12

∴ x – 4y = -12 + 3

∴ x – 4y = -9 …(ii)

Subtracting equation (ii) from (i), we get

∴ x = 31 – 8

∴ x = 23

The present ages of Manish and Savita are 23 years and 8 years respectively.

v. In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.

Solution:

Let the daily wages of skilled workers be ₹ x

that of unskilled workers be ₹ y.

According to the first condition,

∴ The daily wages of skilled workers is ₹ 450 and that of unskilled workers is ₹ 270.

vi. Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A), Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.

Solution:

Let the speeds of Hamid and Joseph be x km/hr andy km/hr respectively.

Distance travelled by Hamid in 20 minutes

According to the first condition,

∴ The speeds of Hamid and Joseph 50 km/hr and 40 km/hr respectively.